Let $(M,g)$ be a Riemannian manifold. The musical isomorphisms $^\flat:\chi(M) \to \Omega^1(M)$ and $^\sharp:\Omega^1(M) \to \chi(M)$ allow the space of differential one-forms $\Omega^1(M)$ to be identified with the space of vector fields $\chi(M)$.

If I'm not mistaken, I can define the Lie bracket of two differential one forms $\alpha,\beta$ by $$[\alpha,\beta] := [\alpha^\sharp, \beta^\sharp]^\flat.$$

Now suppose that $\alpha$ and $\beta$ are exact; i.e. there exist smooth functions $A,B:M\to \mathbb{R}$ such that $\alpha = dA$ and $\beta = dB$, where $``d"$ denotes the exterior derivative. Is it necessarily true that $[dA,dB] = 0$?

Here is the motivation for my question: On a $k$-manifold $M$, the integral curves of $k$ vector fields linearly independent at the point $x \in M$ are the coordinate curves of a local coordinate system centered at $x$ if and only if their pairwise Lie brackets are zero (see, e.g., the discussion here). On a Riemannian manifold, differential one-forms also have integral curves after identifying these one-forms with vector fields. I would like to know if there are "natural" conditions on a collection of $k$ differential one-forms that determine whether their integral curves similarly form the coordinate lines of a coordinate chart.

  • 2
    $\begingroup$ This is just the Lie bracket of gradient vector fields. Try some examples on $\mathbb R^2$. $\endgroup$ Mar 16 '16 at 4:27
  • $\begingroup$ I should have done this from the beginning. Thank you. $\endgroup$ Mar 16 '16 at 11:43

Answer: The answer to my question is that $[dA,dB]$ is not necessarily zero. As a counterexample, consider $\mathbb{R}^2$ with the standard inner product and take $A(x,y) = x^2y^2$, $B(x,y) = xy^3$ (I selected these arbitrarily without thinking about them). Then $\nabla A(x,y) = (dA)^\sharp(x,y) = 2xy^2\frac{\partial}{\partial x} + 2x^2y \frac{\partial}{\partial y}$ and $\nabla B(x,y) = (dB)^\sharp(x,y) = y^3 \frac{\partial}{\partial x} + 3xy^2\frac{\partial}{\partial y}$.

Computing the Lie Bracket of $\nabla A$ and $\nabla B$ in coordinates shows that $[dA,dB](x,y) = [\nabla A, \nabla B](x,y) = (-6x^2y^3 - 2y^5)\frac{\partial}{\partial x} + (2xy^4 + 6x^3y^2)\frac{\partial}{\partial y}$ which is not zero at all points.

Resolving my misunderstanding: The following is an explanation of the misunderstanding which prompted my question. Let $A_1,\ldots,A_k$ be functions on a Riemannian $k$-manifold $(M,g)$ such that the collection of vectors $\{\nabla A_i := (dA_i)^\sharp\}$ are linearly independent at the point $x_0 \in M$. This is true if and only if the collection of forms $\{dA_1,\ldots,dA_k\}$ are themselves linearly independent as linear functionals at $x_0 \in M$. Then the derivative of the map $\varphi: M \to \mathbb{R}^k$, $\varphi: x \mapsto (A_1(x),\ldots,A_k(x))$ is full rank at $x_0 \in M$, so the inverse function theorem guarantees that there exists an open set $U \ni x$ such that $\varphi|_U: U \to \varphi(U)$ is a diffeomorphism. Initially, I thought that the pairwise Lie brackets of all of the $\nabla A_i$ must therefore be zero. However, this is only necessarily be true if the integral curves of the $\nabla A_i$ were the coordinate lines determined by the chart $\varphi$. The definition of $\nabla A_i$ relies on the particular metric $g$, and in general $g$ has nothing to do with $\varphi$; the coordinate lines of $\varphi$ which have nothing do with $g$ are thus not the integral curves of the $\nabla A_i$ in general.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.